1 条题解

  • 0
    @ 2026-10-27 4:21:39

    思路与算法

    将道路按造价递增排序,用并查集维护当前连通块;只在两端属于不同块时选边并合并。输出选出的n-1条边的两个城市编号。造价使用精确大整数。

    正确性说明

    已选森林与任意最优树比较,下一条最便宜的跨块边跨过某个割;若最优树不含它,可用它替换树上跨割边,费用不增加。重复交换保证所选边可扩展为最优树。输入连通,合并n-1次得到覆盖所有城市的无环连通图,即最小生成树。

    复杂度

    O(e log e D+e alpha(n))时间、O(eD+n)空间,含边读入、排序和并查集初始化。D为造价位数。

    易错点

    题目要输出方案而非最小费用数字。平行边按最便宜实际边处理,不同最优树都合法;本题采用完整树合法性和独立最小费用特判。

    参考代码(C++11)

    #include <bits/stdc++.h>
    using namespace std;
    struct cpp_int {
        string s;
        bool neg;
        cpp_int(long long v = 0) : s(to_string(v)), neg(v < 0) {
            if (neg)
                s.erase(s.begin());
        }
        void norm() {
            auto p = s.find_first_not_of('0');
            s = p == string::npos ? "0" : s.substr(p);
            if (s == "0")
                neg = false;
        }
        static int cmpabs(const cpp_int &a, const cpp_int &b) {
            if (a.s.size() != b.s.size())
                return a.s.size() < b.s.size() ? -1 : 1;
            return a.s == b.s ? 0 : a.s < b.s ? -1 : 1;
        }
        friend bool operator<(const cpp_int &a, const cpp_int &b) {
            if (a.neg != b.neg)
                return a.neg;
            int c = cmpabs(a, b);
            return a.neg ? c > 0 : c < 0;
        }
        friend bool operator==(const cpp_int &a, const cpp_int &b) {
            return a.neg == b.neg && a.s == b.s;
        }
        friend bool operator!=(const cpp_int &a, const cpp_int &b) {
            return !(a == b);
        }
        friend bool operator>(const cpp_int &a, const cpp_int &b) {
            return b < a;
        }
        friend bool operator<=(const cpp_int &a, const cpp_int &b) {
            return !(b < a);
        }
        friend bool operator>=(const cpp_int &a, const cpp_int &b) {
            return !(a < b);
        }
        cpp_int operator-() const {
            cpp_int r = *this;
            if (r.s != "0")
                r.neg = !r.neg;
            return r;
        }
        friend cpp_int operator+(const cpp_int &a, const cpp_int &b) {
            cpp_int r;
            r.s.clear();
            int carry = 0;
            if (a.neg == b.neg) {
                int i = (int)a.s.size() - 1, j = (int)b.s.size() - 1;
                while (i >= 0 || j >= 0 || carry) {
                    int v = carry + (i >= 0 ? a.s[i--] - '0' : 0) + (j >= 0 ? b.s[j--] - '0' : 0);
                    r.s += char('0' + v % 10);
                    carry = v / 10;
                }
                r.neg = a.neg;
            } else {
                const cpp_int *x = &a, *y = &b;
                if (cmpabs(a, b) < 0)
                    swap(x, y);
                int i = (int)x->s.size() - 1, j = (int)y->s.size() - 1;
                while (i >= 0) {
                    int v = x->s[i--] - '0' - carry - (j >= 0 ? y->s[j--] - '0' : 0);
                    carry = v < 0;
                    if (carry)
                        v += 10;
                    r.s += char('0' + v);
                }
                r.neg = x->neg;
            }
            reverse(r.s.begin(), r.s.end());
            r.norm();
            return r;
        }
        friend cpp_int operator-(const cpp_int &a, const cpp_int &b) {
            return a + (-b);
        }
        friend cpp_int operator*(const cpp_int &a, int b) {
            cpp_int r;
            r.s.clear();
            long long carry = 0;
            long long z = b;
            if (z < 0)
                z = -z;
            for (int i = (int)a.s.size() - 1; i >= 0; i--) {
                carry += (a.s[i] - '0') * z;
                r.s += char('0' + carry % 10);
                carry /= 10;
            }
            while (carry) {
                r.s += char('0' + carry % 10);
                carry /= 10;
            }
            reverse(r.s.begin(), r.s.end());
            r.neg = a.neg ^ (b < 0);
            r.norm();
            return r;
        }
        friend cpp_int operator/(const cpp_int &a, int b) {
            cpp_int r;
            r.s.clear();
            long long rem = 0, z = b;
            if (z < 0)
                z = -z;
            for (char c : a.s) {
                rem = rem * 10 + c - '0';
                r.s += char('0' + rem / z);
                rem %= z;
            }
            r.neg = a.neg ^ (b < 0);
            r.norm();
            return r;
        }
        cpp_int &operator+=(const cpp_int &b) {
            *this = *this + b;
            return *this;
        }
        cpp_int &operator-=(const cpp_int &b) {
            *this = *this - b;
            return *this;
        }
        cpp_int &operator*=(int b) {
            *this = *this * b;
            return *this;
        }
        friend istream &operator>>(istream &in, cpp_int &a) {
            string v;
            if (in >> v) {
                a.neg = v[0] == '-';
                if (v[0] == '-' || v[0] == '+')
                    v.erase(v.begin());
                a.s = v;
                a.norm();
            }
            return in;
        }
        friend ostream &operator<<(ostream &out, const cpp_int &a) {
            if (a.neg)
                out << '-';
            return out << a.s;
        }
    };
    struct Edge {
        int u, v;
        cpp_int w;
    };
    int main() {
        ios::sync_with_stdio(false);
        cin.tie(nullptr);
        int n, m;
        cin >> n >> m;
        vector<Edge> e(m);
        for (auto &x : e)
            cin >> x.u >> x.v >> x.w;
        sort(e.begin(), e.end(), [](const Edge &a, const Edge &b) {
            if (a.w != b.w)
                return a.w < b.w;
            return tie(a.u, a.v) < tie(b.u, b.v);
        });
        vector<int> p(n + 1);
        iota(p.begin(), p.end(), 0);
        function<int(int)> find = [&](int u) {
            return p[u] == u ? u : p[u] = find(p[u]);
        };
        for (auto x : e)
            if (find(x.u) != find(x.v)) {
                p[find(x.u)] = find(x.v);
                cout << x.u << " " << x.v << "\n";
            }
    }
    
    • 1

    信息

    ID
    1987
    时间
    1000ms
    内存
    64MiB
    难度
    普及−
    标签
    递交数
    0
    已通过
    0
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